# Science:Math Exam Resources/Courses/MATH152/April 2010/Question B 05 (b)/Solution 1

Continuing from the hint, to choose real-valued solutions, we choose ${\displaystyle C_{2}={\bar {C_{1}}}}$, so that the general solution becomes:

${\displaystyle \mathbf {y} =C_{1}\mathbf {y_{1}} +C_{2}\mathbf {y_{2}} =C_{1}\mathbf {y_{1}} +{\overline {C_{1}\mathbf {y_{1}} }}=2\mathrm {Re} \left[C_{1}\mathbf {y_{1}} \right]}$

Recall that

${\displaystyle \mathbf {y_{1}} =\mathbf {k_{1}} \exp {\lambda _{1}t}=\left[{\begin{array}{c}2+3i\\1\end{array}}\right]\exp {(1+2i)t}}$

Since

${\displaystyle \mathrm {Re} \left[\mathbf {y_{1}} \right]=\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}-3\\0\end{array}}\right]e^{t}\sin {2t}}$
${\displaystyle \mathrm {Im} \left[\mathbf {y_{1}} \right]=\left[{\begin{array}{c}3\\0\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\sin {2t}}$

The general form of real-valued solutions is:

{\displaystyle {\begin{aligned}\mathbf {y} &=D_{1}\mathrm {Re} \left[\mathbf {y_{1}} \right]+D_{2}\mathrm {Im} \left[\mathbf {y_{1}} \right]\\&=D_{1}\left(\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}-3\\0\end{array}}\right]e^{t}\sin {2t}\right)\\&+D_{2}\left(\left[{\begin{array}{c}3\\0\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\sin {2t}\right)\end{aligned}}}

where ${\displaystyle D_{1}}$ and ${\displaystyle D_{2}}$ are real constants.

Their relationship with the previous constants are ${\displaystyle D_{1}=2\mathrm {Re} \left[C_{1}\right]}$ and ${\displaystyle D_{2}=-2\mathrm {Im} \left[C_{1}\right]=e^{t}}$

Now if we impose the initial condition ${\displaystyle \mathbf {y} (0)=\left[{\begin{array}{c}1\\0\end{array}}\right]}$, we get

${\displaystyle D_{1}\left[{\begin{array}{c}2\\1\end{array}}\right]+D_{2}\left[{\begin{array}{c}3\\0\end{array}}\right]=\left[{\begin{array}{c}1\\0\end{array}}\right]}$

Thus, ${\displaystyle D_{1}=0~}$ and ${\displaystyle D_{2}={\frac {1}{3}}}$, and the solution of the initial value problem is

${\displaystyle {\frac {1}{3}}\left(\left[{\begin{array}{c}3\\0\end{array}}\right]e^{t}\cos {2t}+\left[{\begin{array}{c}2\\1\end{array}}\right]e^{t}\sin {2t}\right)={\frac {e^{t}}{3}}\left[{\begin{array}{c}3\cos {2t}+2\sin {2t}\\\sin {2t}\end{array}}\right]}$

Alternatively, one can solve directly from

${\displaystyle \mathbf {y} =2\mathrm {Re} \left[C_{1}\mathbf {y_{1}} \right]}$

And impose the initial condition ${\displaystyle \mathbf {y} _{0}=\left[{\begin{array}{c}1\\0\end{array}}\right]}$ direcly to find that ${\displaystyle C_{1}=-{\frac {i}{6}}}$.

Then compute the real part to get the same result as above. The amount of algebra involved will be slightly less.